JEE Main20246 Apr 2024Evening ShiftPhysicsCurrent ElectricityActual
The number of electrons flowing per second in the filament of a 110 ~W bulb operating at 220 ~V is : ( . Given . e =1.6 10⁻¹⁹ C )
Options
- A6.25 10¹⁷
- B1.25 10¹⁹
- C6.25 10¹⁸
- D31.25 10¹⁷
Correct answer
D. 31.25 10¹⁷
Step-by-step solution
Power ( P )= V . I aligned & 110=(220)( I ) & I =0.5 ~A aligned Now, I= n e t aligned & 0.5= ( n t ) (1.6 10⁻¹⁹ ) & n t = 0.5 1.6 10⁻¹⁹ & n t =31.25 10¹⁷ aligned